I am looking for an equivalent of the following integral: $$\int_{0}^{\pi}\frac {\sin x}{1+\cos²(nx)}\mathrm dx$$ when $n\to+\infty$.
Any hint or solution will be welcome. Thanks in advance
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$\begingroup$ What do you think is the limit? What if the numerator is simpler, e.g. equal to 1? What if the numerator is 1 and there is a general integration interval $[a,b]$? Does that give you an idea? $\endgroup$– Hans EnglerCommented Apr 5, 2019 at 18:24
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$\begingroup$ i think that the limit is 0, "if u=nx; i find $\int_0^{n\pi} {\frac{sin(\frac{u}{n})}{1+{cos({u})}^2}}du$" $\endgroup$– PascalCommented Apr 5, 2019 at 18:43
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$\begingroup$ While $sin(u/n)\rightarrow 0$, a t the same time the upper limit of the integral tends to infinity; I don't thhink you can take the limit in such case. What I think is that because $\cos(nx)$ is oscillating very fast, then in the limit function $1/(1+\cos^2(nx))$ can be substituded by its average: $$ \frac{1}{\pi} \int_0^\pi\frac{dx}{1+\cos^2(nx)} = \frac{1}{\sqrt{2}}$$ So I expect that $$ \int_0^\pi\frac{\sin x\,dx}{1+\cos^2(nx)} = \frac{1}{\sqrt{2}}\int_0^\pi\sin x\,dx = \sqrt{2}$$ I'll think about a rigorous proof. $\endgroup$– Adam LatosińskiCommented Apr 5, 2019 at 18:54
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$\begingroup$ What do you call "an equivalent" ?? $\endgroup$– user65203Commented Apr 5, 2019 at 19:02
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$\begingroup$ @AdamLatosiński: First prove that $\int_a^b (1 + \cos^2(n x))^{-1} dx \to (b-a)/\sqrt{2}$ for all $a < b$. This extends to step functions by linearity. Then extend it to piecewise continuous functions with a limit argument. $\endgroup$– Hans EnglerCommented Apr 5, 2019 at 19:02
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1 Answer
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If you make the substitution $x=\frac{u}{n}$ you get \begin{align} \int_0^\pi \frac{\sin x}{1+\cos^2(nx)} {\rm d}x &= \int_0^{n\pi} \frac{\sin\frac{u}{n}}{1+\cos^2u}\frac{{\rm d}u}{n} = \sum_{k=0}^{n-1} \int_{k\pi}^{(k+1)\pi} \frac{\sin\frac{u}{n}}{1+\cos^2u}\frac{{\rm d}u}{n} = \\ &= \sum_{k=0}^{n-1} \int_{0}^{\pi} \frac{\sin\frac{k\pi + u}{n}}{1+\cos^2(k\pi+u)}\frac{{\rm d}u}{n} = \\ &= \sum_{k=0}^{n-1} \int_{0}^{\pi} \frac{\sin\frac{k\pi + u}{n}}{1+\cos^2u}\frac{{\rm d}u}{n} = \\ &= \int_{0}^{\pi} \frac{1}{1+\cos^2u} \left(\frac{1}{n}\sum_{k=0}^{n-1} \sin\frac{k\pi + u}{n} \right){\rm d}u =\end{align}
From the definition of Riemann integral, for every $u\in[0,\pi]$ we have $$\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{k=0}^{n-1} \sin\frac{k\pi + u}{n} = \int_0^1 \sin(\pi x){\rm d}x = \frac{2}{\pi}$$ Alternatively, we can note that \begin{align} \frac{1}{n}\sum_{k=0}^{n-1} \sin\frac{k\pi + u}{n} & = \frac{1}{n}\sum_{k=0}^{n-1} \Im\left(\exp\left({\rm i}\frac{k\pi+u}{n}\right)\right) = \\ &= \frac{1}{n} \Im\left(\exp\left({\rm i}u/n\right)\frac{1-\exp({\rm i}\pi)}{1-\exp({\rm i}\pi/n)}\right) = \\ &= \frac{1}{n} \Im\left(\exp\left({\rm i}(u-\pi/2)/n\right)\frac{2}{-2{\rm i} \sin(\pi/(2n))}\right) = \\ &= \frac{1}{n} \frac{\cos\left((u-\pi/2)/n\right)}{\sin(\pi/(2n))} \rightarrow^{n\rightarrow\infty} \frac{1}{\pi/2} =\frac{2}{\pi} \end{align} Moreover, this is an uniform convergence, so we can get with the limit inside the integral: \begin{align} & \lim_{n\rightarrow\infty}\int_0^\pi \frac{\sin x}{1+\cos^2(nx)} {\rm d}x = \\ & = \lim_{n\rightarrow\infty}\int_{0}^{\pi} \frac{1}{1+\cos^2u} \left(\frac{1}{n}\sum_{k=0}^{n-1} \sin\frac{k\pi + u}{n} \right){\rm d}u = \\ & = \int_{0}^{\pi} \frac{1}{1+\cos^2u} \lim_{n\rightarrow\infty}\left(\frac{1}{n}\sum_{k=0}^{n-1} \sin\frac{k\pi + u}{n} \right){\rm d}u =\\ & = \int_{0}^{\pi} \frac{1}{1+\cos^2u} \frac{2}{\pi} {\rm d}u = \sqrt{2}\end{align}
